Question -
Answer -
(i)┬а2 + 5 + 8 + тАж + 182
First term, a = a1┬а=2
Common difference, d =a2┬атАУ a1┬а= 5 тАУ 2 = 3
an┬аtermof given AP is 182
an┬а= a+ (n-1) d
182 = 2 + (n-1) 3
182 = 2 + 3n тАУ 3
182 = 3n тАУ 1
3n = 182 + 1
n = 183/3
= 61
Now,
By using the formula,
S = n/2 (a + l)
= 61/2 (2 + 182)
= 61/2 (184)
= 61 (92)
= 5612
тИ┤ The sum of the seriesis 5612
(ii)┬а101 + 99 + 97 + тАж + 47
First term, a = a1┬а=101
Common difference, d =a2┬атАУ a1┬а= 99 тАУ 101 = -2
an┬аtermof given AP is 47
an┬а= a+ (n-1) d
47 = 101 + (n-1)(-2)
47 = 101 тАУ 2n + 2
2n = 103 тАУ 47
2n = 56
n = 56/2 = 28
Then,
S = n/2 (a + l)
= 28/2 (101 + 47)
= 28/2 (148)
= 14 (148)
= 2072
тИ┤ The sum of the seriesis 2072
(iii)┬а(a тАУ b)2┬а+(a2┬а+ b2) + (a + b)2┬а+ sтАж. + [(a +b)2┬а+ 6ab]
First term, a = a1┬а=(a-b)2
Common difference, d =a2┬атАУ a1┬а= (a2┬а+ b2)тАУ (a тАУ b)2┬а= 2ab
an┬аtermof given AP is [(a + b)2┬а+ 6ab]
an┬а= a+ (n-1) d
[(a +b)2┬а+ 6ab] = (a-b)2┬а+ (n-1)2ab
a2┬а+ b2┬а+2ab + 6ab = a2┬а+ b2┬атАУ 2ab + 2abn тАУ 2ab
a2┬а+ b2┬а+8ab тАУ a2┬атАУ b2┬а+ 2ab + 2ab = 2abn
12ab = 2abn
n = 12ab / 2ab
= 6
Then,
S = n/2 (a + l)
= 6/2 ((a-b)2┬а+[(a + b)2┬а+ 6ab])
= 3 (a2┬а+b2┬атАУ 2ab + a2┬а+ b2┬а+ 2ab + 6ab)
= 3 (2a2┬а+2b2┬а+ 6ab)
= 3 ├Ч 2 (a2┬а+b2┬а+ 3ab)
= 6 (a2┬а+b2┬а+ 3ab)
тИ┤ The sum of the seriesis 6 (a2┬а+ b2┬а+ 3ab)